What is the last digit of: 3^2046?
\(\begin{array}{|rcll|} \hline && 3^{\varphi(10) } \equiv 1 \pmod{10} \quad | \quad \varphi(10) = 10\cdot(1-\frac12)\cdot(1-\frac15) = 4 \\ && 3^4 \equiv 1 \pmod{10} \\\\ && 3^{2046} \pmod{10} \\ &\equiv & 3^{4\cdot 511 +2} \pmod{10} \\ &\equiv & 3^{4\cdot 511}\cdot 3^2 \pmod{10} \\ &\equiv & (3^4)^{511}\cdot 3^2 \pmod{10} \\ &\equiv & (1)^{511}\cdot 3^2 \pmod{10} \\ &\equiv & 3^2 \pmod{10} \\ &\equiv & 9 \pmod{10} \\ &\equiv & 9 \\ \hline \end{array} \)
The last digit is 9
3^1 = 3
3^2 = 9
3^3 = 27
3^4 =81
3^5 = 243
3^6 = 729
3^7 =2187
3^8 = 6561
3^9 =19683
3^10=59049
You might be able to see a pattern of last digits happening here? (repeating every four?)
2046 mod 4 = 2 The second one above is 9 I expect 3^2046 ends in 9.
What is the last digit of: 3^2046?
\(\begin{array}{|rcll|} \hline && 3^{\varphi(10) } \equiv 1 \pmod{10} \quad | \quad \varphi(10) = 10\cdot(1-\frac12)\cdot(1-\frac15) = 4 \\ && 3^4 \equiv 1 \pmod{10} \\\\ && 3^{2046} \pmod{10} \\ &\equiv & 3^{4\cdot 511 +2} \pmod{10} \\ &\equiv & 3^{4\cdot 511}\cdot 3^2 \pmod{10} \\ &\equiv & (3^4)^{511}\cdot 3^2 \pmod{10} \\ &\equiv & (1)^{511}\cdot 3^2 \pmod{10} \\ &\equiv & 3^2 \pmod{10} \\ &\equiv & 9 \pmod{10} \\ &\equiv & 9 \\ \hline \end{array} \)
The last digit is 9